Timeline for How to simplify expression depending on former definition?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Dec 22, 2023 at 12:31 | comment | added | user57467 | Lericr, How would I do it then with Hold forms? I want to use different forms by which you can describe triangles (by a,b,c or angles or other parts once coordinates are given) | |
| Dec 21, 2023 at 16:15 | comment | added | lericr | What are you going to do with the 1+b/a expression? If you just want to see it, you can use one of the Hold forms (for example). | |
| Dec 21, 2023 at 15:49 | history | became hot network question | |||
| Dec 21, 2023 at 13:44 | comment | added | user57467 |
I found that the following works eq1 = a == x + y; eq2 = b == x - y; eq3 = u == a + b; eq4 = v == a; eq5 = c == u/v; sol = Eliminate[{eq1, eq2, eq3, eq4, eq5}, {x, y, u, v}]; Solve[sol, c] Which looks very clumsy. Any better ideas?
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| Dec 21, 2023 at 12:48 | review | Close votes | |||
| Jan 8 at 3:06 | |||||
| Dec 21, 2023 at 12:31 | comment | added | MarcoB | Does this answer your question? Can I simplify an expression into form which uses my own definitions? | |
| Dec 21, 2023 at 7:21 | comment | added | user57467 | 1+b/a is ok, I just wanted to avoid long Replace answers. I was looking for some kind of short switch (from x,y to a,b), because I have many variables to deal with. I was looking sore something like Eliminate, but that works only for equations. | |
| Dec 21, 2023 at 6:22 | comment | added | lericr |
Or are you saying you want 1 + (x-y)/(x+y)?
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| Dec 21, 2023 at 6:19 | comment | added | lericr |
Why does it matter if a and b are long expressions if you just want 1+b/a?
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| Dec 21, 2023 at 5:59 | answer | added | bmf | timeline score: 3 | |
| Dec 21, 2023 at 5:55 | history | edited | user64494 | CC BY-SA 4.0 |
edited title
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| Dec 21, 2023 at 5:50 | history | asked | user57467 | CC BY-SA 4.0 |